math185f09-hw5sol

math185f09-hw5sol - MATH 185 COMPLEX ANALYSIS FALL 2009/10...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 5 SOLUTIONS 1. Let a ∈ R and z ∈ C . (a) Evaluate the following integrals Z 1 e it cos atdt and Z 1- 1 dt t 2 + i . Solution. For the first integral, we have Z 1 e it cos atdt = Z 1 cos t cos atdt + i Z 1 sin t cos atdt. Since cos t cos at = 1 2 [cos( t + at ) + cos( t- at )] , sin t sin at = 1 2 [sin( t + at ) + sin( t- at )] , we obtain Z 1 e it cos atdt = 1 2 sin(1 + a ) t 1 + a + sin(1- a ) t 1- a 1- i 2 cos(1 + a ) t 1 + a + cos(1- a ) t 1- a 1 = 1 2 sin(1 + a ) 1 + a + sin(1- a ) 1- a- i 2 cos(1 + a ) 1 + a + cos(1- a ) 1- a- 2 1- a 2 . For the second integral, we first apply the formula for complex square roots in Chapter 1 and then find a partial fraction decomposition. 1 t 2 + i = 1 t- 1 √ 2 + i √ 2 t + 1 √ 2- i √ 2 = α t- 1 √ 2 + i √ 2 + β t + 1 √ 2- i √ 2 and so α = 1 2 √ 2 (1 + i ) , β =- 1 2 √ 2 (1 + i ) . Therefore Z 1- 1 dt t 2 + i = 1 2 √ 2 (1 + i )log t- 1 √ 2 + i √ 2 1- 1- 1 2 √ 2 (1 + i )log t + 1 √ 2- i √ 2 1- 1 . (b) Show that if Re z >- 1, then the integral R 1 t z dt exists and Z 1 t z dt ≤ 1 1 + Re z . Solution. Note that for 0 ≤ t ≤ 1, | t z | = t Re z and so Z 1 t z dt ≤ Z 1 | t z | dt = Z 1 t Re z dt. Date : October 23, 2009 (Version 1.0). 1 Since for Re z >- 1 the last integral converges, R 1 t z dt exists and the required inequality follows from Z 1 t Re z dt = t Re z +1 1 + Re z 1 = 1 1 + Re z . (c) Show that if a < 1, then Z 1- 1 cos it t a dt ≤ 2 Z 1- 1 dt t a and thus the (improper) integral R 1- 1 t- a cos itdt converges absolutely. Solution. Note that cos it = cosh t and for- 1 ≤ t ≤ 1, we have 0 ≤ cosh t < 2. Hence Z 1- 1 cos it t a dt ≤ Z 1- 1 cos it t a dt = Z 1- 1 cosh t | t a | dt ≤ 2 Z 1- 1 dt t a . Since the last integral converges for a < 1, we see that R 1- 1 t- a cos itdt converges absolutely. 2. (a) For k = 1 , 2 , 3, evaluate the following integrals Z Γ k Re( z ) dz, Z Γ k z 2 dz, Z Γ k dz z along the curves from the point z = 1 to z 1 = i in the counter clockwise direction as described in Figure 1. Figure 1. Left : Γ 1 is along the boundary of the square: { x + iy | ≤ x ≤ 1 , ≤ y ≤ 1 } . Center : Γ 2 is along the boundary of the circle: { e it | ≤ t ≤ π/ 2 } . Right : Γ 3 is along the line segment: { (1- t ) + it | ≤ t ≤ 1 } ....
View Full Document

This note was uploaded on 02/17/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at Berkeley.

Page1 / 8

math185f09-hw5sol - MATH 185 COMPLEX ANALYSIS FALL 2009/10...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online